A train starts from rest and accelerates uniformly at a rate of 2 m s^-2 for 10 s. It then maintains a constant speed for 200 s. The brakes are then applied and the train is uniformly retarded and comes to rest in 50 s.

Find —

(i) the maximum velocity reached,

(ii) the retardation in the last 50 s,

(iii) the total distance travelled, and

(iv)the average velocity of the train.

As we know,

v = u + at

Given,

u = 0

t = 10 s

a = 2 m s^-2

Substituting the values in the formula above we get,

$v = 0 + 2 \times 10 \\[0.5em] \Rightarrow v = 20 \\[0.5em]$

Hence, final velocity = 20 m s^-1

(ii) Retardation in last 50 s,

As we know,

v = u + at

Given,

u = 20 m s^-1

v = 0 m s^-1

t = 50 s

Substituting the values in the formula above we get,

$0 = 20 + [(- a) \times (50)] \\[0.5em] \Rightarrow 50 \times a = 20 \\[0.5em] \Rightarrow a = \dfrac{20}{50} \\[0.5em] \Rightarrow a = 0.4 \text{ m s}^{-2} \\[0.5em]$

Hence, retardation in the last 50 s = 0.4 m s^-2

(iii) As we know,

S = ut + $\dfrac{1}{2}$ at²

For the first 10 s,

u = 0

a = 2 ms^-2

$S${1} = (0 \times 10) + (\dfrac{1}{2} \times 2 \times 10^2) \\[0.5em] S{1} = 0 + (\dfrac{1}{2} \times 2 \times 100) \\[0.5em] \Rightarrow S_{1} = 100 \text{ m}

Hence, S₁ = 100 m   …. [1]

For the next 200 s,

S₂ = speed x time

where speed = 20

time = 200 s

Substituting the values in the formula above we get,

$S$2 = 20 \times 200 \\[0.5em] \Rightarrow S2 = 4000 \text{ m} \\[0.5em]

Hence, S₂ = 4000 m   …. [2]

For the next 50 s

$S${3} = (20 \times 50) + [\dfrac{1}{2} \times (- 0.4) \times 50^2] \\[0.5em] S{3} = 1000 - [0.2 \times 2500] \\[0.5em] S{3} = 1000 - [0.2 \times 2500] \\[0.5em] \Rightarrow S{2} = 1000 - 500 \text{ m} \\[0.5em] \Rightarrow S_{2} = 500 \text{ m}

Hence, S₃ = 500 m   …. [3]

So,

Total distance = sum of [1], [2] and [3]

$S = 100 \text{ m} + 4000 \text{ m} + 500 \text{ m} \\[0.5em] \Rightarrow S = 4600 \text{ m}$

Hence, total distance covered = 4600 m

(iv) average velocity = $\dfrac{\text {total distance covered}}{\text {time taken}}$

Substituting the values in the formula above we get,

$\text {average velocity} = \dfrac{4600}{260} \\[0.5em] \Rightarrow \text {average velocity} = 17.69 \text{ m s}^{-1}$

Hence, average velocity = 17.69 m s^-1